Population Standard Deviation

So, we should reject the null hypothesis H0. At a 0.05 level of significance level, we conclude that there is a significant difference between the average height for females and the average height for the males.

|A random group of apartments was selected from a city to analyze the number of bedrooms they have. Is there evidence to reject |

Use the Internet or Strayer Library to research articles on hypothesis test and its application in business. Select one (1) company or organization which utilized hypothesis test

c. Note the ways in which the means and standard deviations differ, and speculate on the possible meaning of these differences, presuming that they are representative of U.S. governors and large corporations’ CEOs in general.

Results from previous studies showed 79% of all high school seniors from a certain city plan to attend college after graduation. A random sample of 200 high school seniors from this city reveals that 162 plan to attend college. Does this indicate that the percentage has increased from that of previous studies? Test at the 5% level of significance.

 Explain how the data collected will provide the data necessary to support or negate the hypothesis or proposition

I found the average (mean) income to be $43,740, with a standard deviation of $14,640. According to the hypothesis test (see appendix), the calculated test statistic of -3.0236 does fall in the rejection region of z<-1.645 therefore I can reject the null hypothesis and say there is sufficient evidence to indicate µ<50 or $50,000. The p-value of 0.001 (see appendix for data), supports the rejection of the null hypothesis since the p value is less than α=0.05. Based on the confidence interval, we can be 95% certain that the average income lies between $39,680-$47,800.

c) What is the null and alternative hypothesis? Do the data results lead you to reject or fail to reject the null hypothesis?

The sample size in this speculation is greater than 30, therefore, The Central Limit Theorem (CLT) will apply, and no assumptions need to be made.

Place your answer, rounded to 2 decimal places, in the blank. For example, 0.23 is a legitimate entry. 0.35

2. At 95% confidence, use the p-value approach and test to determine if the average yearly income of marketing managers in the East is significantly different from the West.

The customers in this case study have complained that the bottling company provides less than the advertised sixteen ounces of product. They need to determine if there is enough evidence to conclude the soda bottles do not contain sixteen ounces. The sample size of sodas is 30 and has a mean of 14.9. The standard deviation is found to be 0.55. With these calculations and a confidence level of 95%, the confidence interval would be 0.2. There is a 95% certainty that the true population mean falls within the range of 14.7 to 15.1.

Fry Brothers heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of

Conclusion : Reject the null hypothesis. The sample provide enough evidence to support the claim that mean is significantly different from 12 .

Yes, because the statistical value of 9.619 is greater than the probability level of 0.001.